Question

Socially conscious investors screen out stocks of alcohol and tobacco makers, firms with poor environmental records, and companies with poor labor practices. Some examples of "good" socially conscious companies are Johnson and Johnson, Dell Computers, Bank of America, and Home Depot. The question is, are such stocks overpriced? One measure of value is the P/E. or price-to-earnings ratio. High P/E ratios may indicate a stock is overpriced. For the S&P Stock Index of all major stocks, the mean P/E ratio is # 19.4. A random sample of 36 "socially conscious stocks gave a P/E ratio sample mean of x = 17.7, with sample standard deviations 5.2. Does this indicate that the mean P/E ratio of all socially conscious stocks is different (either way) from the mean P/E ratio of the S&P Stock Index? Use a 0.05. USE SALT (a) What is the level of significance? State the null and alternate hypotheses OHO: 19.4; M N = 19.4 O Ho: H = 19.4; Hy! > 19.4 O Hol 19.43 MB<19.4 O HO! 19.4 19.4 O Ho: > 19.4; HTN = 19.4 (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution The Student's t, since the sample size is large and o is known The standard normal, since the sample size is large and ois unknown. The Student's t, since the sample size is large and is unknown. The standard normal, since the sample size is large and is known What is the value of the sample test statistic? (Round your answer to three decimal places)

Answer 1

The provided sample mean is 7 = 17.7 and the sample standard deviation is s = 5.2, and the sample size is n = 36. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: H = 19.4 Ha: u 19.4 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two- tailed test is te = 2.03. The rejection region for this two-tailed test is R={t: \t > 2.03} (3) Test Statistics The t-statistic is computed as follows X - MO t = 17.7 – 19.4 5.2/36 -1.962 8/n (4) Decision about the null hypothesis Since it is observed that t = 1.962 <te = 2.03, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0578, and since p=0.0578 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 19.4, at the 0.05 significance level.

(a)

Level of significance = 0.05

(b)

Correct option : The Student's t , since the sample size is large and $\sigma$ is unknown.